POJ 3160 Father Christmas flymouse 强连通分量+缩点+DP

这道题的大意就是，给出一个有向图，每个点有一个点权，点权可能是正也可能为负，一个人从某点出发，沿着一些路，访问结点，或者仅仅是路过这个结点，而不去访问，最后求他能访问到的最大的点权和。

我们注意到，他对某个结点是可以选择访问或者不访问的，那么只用访问那些点权为正数的点了。

首先，求强连通分量，缩点，然后新点的点权就是原强连通分量中，所有正点权之和。

之后就要进行DP求解，我在本题中使用了Kosaraju算法，其好处就是，最后缩点后，恰好形成拓扑序列，而DP就要从拓扑序靠后的点往前进行DP，这样才能从叶子结点往根部进行DP，从而得到正确答案

/*
ID: sdj22251
PROG: subset
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define LOCA
#define MAXN 5005
#define INF 100000000
#define eps 1e-7
using namespace std;
struct Edge
{
int v, next;
}edge[10 * MAXN], revedge[10 * MAXN], newedge[MAXN];
int head[MAXN], revhead[MAXN], e, visited[MAXN], newhead[MAXN];
int order[MAXN], cnt, id[MAXN], val[MAXN], newval[MAXN];
int uu[5 * MAXN], vv[5 * MAXN], out[MAXN], dp[MAXN];
int n, m;
int ans;
void init()
{
e = 0;
memset(head, -1, sizeof(head));
memset(revhead, -1, sizeof(revhead));
memset(newhead, -1, sizeof(newhead));
memset(out, 0 , sizeof(out));
memset(newval, 0, sizeof(newval));
memset(dp, 0, sizeof(dp));
}
void insert(const int &x, const int &y)
{
edge[e].v = y;
edge[e].next = head[x];
head[x] = e;
revedge[e].v = x;
revedge[e].next = revhead[y];
revhead[y] = e;
e++;
}
void newinsert(const int &x, const int &y)
{
newedge[e].v = y;
newedge[e].next = newhead[x];
newhead[x] = e++;
}
void readdata()
{
for(int i = 1; i <= n; i++)
scanf("%d", &val[i]);
for(int i = 0; i < m; i++)
{
scanf("%d%d", &uu[i], &vv[i]);
uu[i]++;
vv[i]++;
insert(uu[i], vv[i]);
}
}
void dfs(int u)
{
visited[u] = 1;
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].v;
if(!visited[v])
dfs(v);
}
order[cnt++] = u;
}
void dfs_rev(int u)
{
visited[u] = 1;
id[u] = cnt;
if(val[u] > 0) newval[cnt] += val[u];
for(int i = revhead[u]; i != -1; i = revedge[i].next)
{
int v = revedge[i].v;
if(!visited[v])
dfs_rev(v);
}
}
void Kosaraju()
{
init();
readdata();
memset(visited, 0, sizeof(visited));
cnt = 0;
for(int i = 1; i <= n; i++)
{
if(!visited[i])
dfs(i);
}
memset(visited, 0, sizeof(visited));
cnt = 0;
for(int i = n - 1; i >= 0; i--)
{
if(!visited[order[i]])
{
cnt++;
dfs_rev(order[i]);
}
}
e = 0;
for(int i = 0; i < m; i++)
{
int u = id[uu[i]];
int v = id[vv[i]];
if(u != v) newinsert(u, v);
}
ans = 0;
for(int u = cnt; u >= 1; u--)
{
int tmp = 0;
dp[u] = newval[u];
for(int i = newhead[u]; i != -1; i = newedge[i].next)
{
int v = newedge[i].v;
tmp = max(tmp, dp[v]);
}
dp[u] += tmp;
ans = max(ans, dp[u]);
}
}
int main()
{
while(scanf("%d%d", &n, &m) != EOF)
{
Kosaraju();
printf("%d\n", ans);
}
return 0;
}